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This is somewhat similar to my other question here.

Consider the two ellipses given by the equations

\begin{equation} \frac{x^2}{2^2} + \frac{(y-1)^2}{1^2} = 1 \end{equation}

and

\begin{equation} \frac{x^2}{1^2} + \frac{(y-4)^2}{(1/2)^2} = 1. \end{equation}

How do I find the coordinates for the two tangent points of their common tangent at the left side of the ellipses? (I hope the question makes sense.)

Community
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Svend Tveskæg
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2 Answers2

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Let's introduce the following new variables: $$2u=x\ \text{ and }\ y-1=v.$$

With these new variables, we have

$$u^2+v^2=1\ \text{ and }\ u^2+(v-3)^2=\frac14$$

that is, we have two circles as shown in the figure below.

enter image description here

We have similar triangles and we can see that $OD=6$. Also, by the Pythagorean theorem $DC=\frac{\sqrt{35}}2$ and by the similarity of $OBD$ and $O'CD$: $OB=\frac6{\sqrt{35}}$. So, the slope of the red straight lien is $-\sqrt{35}$. Don't forget a about the other tangent, not shown, whose slope is $\sqrt{35}$.

The two tangent lines in the $u,v$ system are

$$v=-\sqrt{35}u+6\ \text{ and } \ v=\sqrt{35}u+6.$$

Returning to the $x,y$ coordinate system, we get

$$y=-\frac{\sqrt{35}}2x+7\ \text{ and } \ y=\frac{\sqrt{35}}2x+7.$$

EDIT

Unforgivable! I forgave the other pair of tangents:

enter image description here

After similar calculations we get the equations of the other pair of tangent lines.

$$y=-\frac{\sqrt3}{2}x+3\ \text{ and }y=\frac{\sqrt3}{2}x+3 \ $$

zoli
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    It’s worth pointing out that this solution works because the ellipses are homothetic. – amd Apr 07 '17 at 19:29
  • In $O'CD$: $\frac14+X^2=9$. From here $X^2=9-\frac14=\frac{36}4-\frac14$. $O'D$ is the hypotenuse. – zoli Apr 07 '17 at 19:39
  • No problemo. (You cannot be ...er than I am.) – zoli Apr 07 '17 at 19:41
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    @SvendTveskæg: I had to edit my answer. – zoli Apr 08 '17 at 12:06
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    @SvendTveskæg: Thank you, but I had one more mistake. Take a look at the equations of the blue tangents. – zoli Apr 09 '17 at 07:16
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    $OBD$ and $O'CD$ are similar because their angles equal. At $O$ and $C$ they have right angles and $D$ is a common angle. The third angles have to be equal because the sum of the angles is $180^{\circ}$. – zoli Apr 11 '17 at 20:35
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    @SvendTveskæg: Not a stupid question. I was wrong. Of course the right result is $\pm \sqrt{3}/2$. – zoli Apr 12 '17 at 10:24
  • @SvendTveskæg: Good to hear. Now, I understand as well. : ) – zoli Apr 12 '17 at 10:26
  • Hi zoil. Am afraid that I have to accept David Quinn's answer instead; it's a more general approach and I finally understood it. Sorry, but thank you very much for the elegant solution! – Svend Tveskæg Apr 18 '17 at 18:24
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Let such a tangent line be $y=mx+c$

Solving simultaneously with the ellipses gives two quadratics in $x$ which are $$(4m^2+1)x^2+8mx(c-1)+4c^2-8c=0$$ and $$(4m^2+1)x^2+8mx(c-4)+4c^2-32c+63=0$$

Each of these must have double roots at the point of tangency and therefore the discriminant is zero, so this leads to two equations in $m$ and $c$ which are $$64m^2(c-1)^2=4(4m^2+1)(4c^2-8c)$$ and $$64m^2(c-4)^2=4(4m^2+1)(4c^2-32c+63)$$

Dividing these eliminates the $m$ terms and we get an equation for $c$ which is $$3c^2-30c+63=0\implies c=3,7$$

A diagram will confirm that the value $c=3$ gives us the tangents which pass between the ellipses, so we choose $c=7$ from which we get the gradient for the common tangent on the left as $$m=+\frac{\sqrt{35}}{2}$$

Using "$x=-\frac{B}{2A}$" the $x$-coordinates of the points of tangency which are $$x_1=-\frac{\sqrt{35}}{3}$$ and $$x_2=-\frac{\sqrt{35}}{6}$$

David Quinn
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  • This looks really nice! Can you please extend your explanations, since I don't understand them to be honest? :-) – Svend Tveskæg Apr 07 '17 at 18:13
  • I have added some extra explanations, but have not included all the algebra. Please let me know if there is anything you don't understand. – David Quinn Apr 07 '17 at 18:55
  • Hmmm. If you could add "all the algebra", I would really appreciate it. For example, how do you come up with the first and second equations (and the third, fourth, and fifth)? I'll gladly accept your answer when I have enough information to understand your solution. :-) – Svend Tveskæg Apr 07 '17 at 19:05
  • The algebra is not difficult - you should try it. But it is time-consuming to put it all in MathJax... – David Quinn Apr 07 '17 at 19:18
  • Can I persuade you to derive the two quadratics equations in $x$ in detail? I hope that I can solve the rest of the problems myself afterwards. – Svend Tveskæg Apr 14 '17 at 21:00
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    You have accepted another answer, so I don't see why you need me to elaborate further on mine. But since you ask, the first quadratic equation is derived from multiplying out and simplifying $$\frac{x^2}{4}+(mx+c-1)^2=1$$ which is quicker for you to do yourself on paper than it is for me to type out in MathJax. :) – David Quinn Apr 14 '17 at 21:15
  • Thank you. I'll look at it later. (I accepted the other answer because I found something that worked and that I understood.) I'm not sure, but does any of your algebra rely on the fact that the two ellipses have common $x$-coordinate as centre? Maybe the fact that the equations have double root at tangency? – Svend Tveskæg Apr 14 '17 at 21:35
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    No the method is independent of the actual ellipse equations. When a line touches a curve there must be a double root... – David Quinn Apr 14 '17 at 21:42
  • I almost got it now. Only remaining question: What is "$x = -\frac{B}{2A}$"? – Svend Tveskæg Apr 15 '17 at 00:48
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    It's the solution of $Ax^2+Bx+C=0$ when $B^2-4AC=0$ – David Quinn Apr 15 '17 at 05:29
  • I think that I've finally put together a small piece if Mathematica code that can calculate all eight tangent points, regardless of the major and minor axes and centres of either ellipses. Thank you very much for your help! – Svend Tveskæg Apr 18 '17 at 18:22
  • You are welcome. Glad to be of help – David Quinn Apr 18 '17 at 18:38